Define (a) heat of formation, (b) heat of combustion, (c) heat of neutralisation - Leaving Cert Chemistry - Question 5 - 2001

Heat of combustion is defined as the heat change that occurs when one mole of a substance is burned completely in excess oxygen, producing products in their standard states.

Heat of neutralisation is defined as the heat change that occurs when one mole of acid reacts with one mole of base to produce one mole of water, usually measured under standard conditions.

The heats of formation of carbon dioxide and water can be determined experimentally because they can be formed directly from their elements under standard conditions. In contrast, ethanoic acid cannot be formed directly from its elements in a simple reaction. Instead, it must undergo complex processes, making it difficult to measure its heat of formation directly.

To calculate the heat of formation of ethanoic acid, we can use Hess's Law. The reaction can be expressed as follows:

To measure the heat of neutralisation, one can use a calorimeter. The procedure involves the following steps:

1. Measure a known volume of ethanoic acid solution in a suitable container (such as a polystyrene cup).
2. Measure an equal volume of sodium hydroxide solution in a separate container.
3. Record the initial temperatures of both solutions.
4. Quickly mix the solutions and record the final temperature once the reaction is complete.
5. Calculate the heat released using the formula: $q = mc\Delta T$ where m is the mass of the solution, c is the specific heat capacity, and $abla T$ is the change in temperature.

The heat of neutralisation of ethanoic acid can be calculated using the given data:

1. Amount of heat produced for ethanoic acid: 27.9 kJ for neutralising 50 cm³ of 1.0 mol dm⁻³.

To find the heat per mole:

$ext{Moles of } ext{C}_2 ext{H}_4 ext{O}_2 = 0.050 ext{ L} imes 1.0 ext{ mol L}^{-1} = 0.050 ext{ mol}$

$ext{Heat of neutralisation} = \frac{27.9 ext{ kJ}}{0.050 ext{ mol}} = 558 ext{ kJ mol}^{-1}$

2. For the sulfuric acid neutralisation:

• 57.2 kJ for the same volume:

  $ext{Moles of H}_2 ext{SO}_4 = 0.050 imes 1.0 = 0.050 ext{ mol}$

  $ext{Heat of neutralisation of H}_2 ext{SO}_4 = \frac{57.2 ext{ kJ}}{0.050 ext{ mol}} = 1144 ext{ kJ mol}^{-1}$

Difference in heats of neutralisation: 1144 - 558 = 586 kJ mol⁻¹